ma’ruza 14. differensial tenglamalarni yechish

DOCX 7 pages 97,4 KB Free download

7 pages

FaylHub.uz

Telegram orqali yuklash

Hajmi 97,4 KB

Fayl turi DOCX

Pages 7

Updated Dek 2025

Page preview (5 pages)

Scroll down 👇

1 / 7

ma’ruza 14. differensial tenglamalarni yechish. reja: 1. dsolve(t,f,options) buyrug’i. 2. yechimning fundamental (bazis) sistemasi. 3. differensial tenglamalarni sonli yechish. tayanch iboralar: differensial tenglama, koshi masalasi, chegaraviy masala. maple muhitida differensial tenglamalarni analitik yechish uchun dsolve(t,f,options) buyrug’i ishlatiladi, bu yerda t – differensial tenglama, f – noma’lum funksiya, options – parametrlar. parametrlar yechish metodlarini ko’rsatadi, masalan, jimlikda analitik yechim quyidagicha izlanadi: type=exact. differensial tenglamalarni tuzishda hosilalarni belgilash uchun diff buyrug’i ishlatiladi, masalan, y''+y=x differensial tenglama quyidagicha yoziladi: diff(y(x),x$2)+y(x)=x. differensial tenglamalarning umumiy yechimi ixtiyoriy o’zgarmasdan, ya’ni differensial tenglama tartibini bildiruvchidan sondan bog’liq bo’ladi. maple da bunday o’zgarmaslar, odatda , _s1, _s2, va hokazo ko’rinishlarda belgilanadi dsolve buyrug’i differensial tenglamalar yechimini hisoblanmaydigan formatda chiqarishni amalga oshiradi. yechim bilan keyinchalik ishlash kerak bo’lsa, (masalan , yechimni grafigini qurish kerak bo’lsa) olingan yechimning chap tomonini rhs(%) buyrug’i bilan ajratish kerak bo’ladi. misollar 1 y'+ycosx=sinx cosx differensial tenglamaning umumiy yechimini toping. > restart; > de:=diff(y(x),x)+y(x)*cos(x)=sin(x)*cos(x); > …

2 / 7

ratadi. buning uchun dsolve buyrug’i parametrida output=basis deb ko’rsatish kerak. misol 1. diffrensial tenglama yechimining fundamental sistemasini toping: y(4)+2y''+y=0. > de:=diff(y(x),x$4)+2*diff(y(x),x$2)+y(x)=0; > dsolve(de, y(x), output=basis); koshi masalasi yoki chegaraviy masalani yechish. dsolve buyrug’i koshi masalasi yoki chegaraviy masalani yechadi, agar differensial tenglama bilan birga noma’lum funksiya uchun boshlang’ich yoki chegaraviy shartlar qo’yilgan bo’lsa. boshlang’ich yoki chegaraviy shartlarda hosilani belgilash uchun differensial operator ishlatiladi, masalan, y''(0)=2 shartni quyidagicha yozish kerak bo’ladi : , yoki y'(1)=0 shart quyidagicha yoziladi: . eslatib qtamizki, n- tartibli hosila ko’rinishda yoziladi. misollar 1. koshi masalasi yechimini toping: y(4)+y''=2cosx, y(0)=- 2, y'(0)=1, y''(0)=0, y'''(0)=0. > de:=diff(y(x),x$4)+diff(y(x),x$2)=2*cos(x); > cond:=y(0)=-2, d(y)(0)=1, (d@@2)(y)(0)=0, (d@@3)(y)(0)=0; cond:=y(0)=- 2, d(y)(0)=1, (d(2))(y)(0)=0, (d(3))(y)(0)=0 > dsolve({de,cond},y(x)); y(x)=- 2cos(x)- xsin(x)+x 2. chegaraviy masalani yeching: , , . yechim grafigini yasang. > restart; de:=diff(y(x),x$2)+y(x)=2*x-pi; > cond:=y(0)=0,y(pi/2)=0; > dsolve({de,cond},y(x)); izoh: yechimni grafigini yasash uchun olingan ifodaning o’ng tomonini ajratish kerak bo’ladi. > y1:=rhs(%):plot(y1,x=-10..20,thickness=2); differensial tenglamalar sistemasi dsolve buyrug’i …

3 / 7

rdagi differensial tenglamalarning aniq analitik yechimini topish qiyin. bunday holda differensial tenglamani taqribiy metodlar orqali yechish mumkin, xususan, noma’lum funksiyani darajali qatorga yoyish orqali. differensial tenglamaning yechimini darajali qator ko’rinishida yechish uchun dsolve buyrug’ida o’zgaruvchidan keyin type=series (yoki oddiy series) parametrni ko’rsatish kerak. qator yoyish darajasi n, ya’ni yoyish amalga oshiriladigan daraja ko’rsatkichini ko’rsatish uchun, dsolve buyrug’ini oldiga daraja tartibini aniqlash buyrug’i order:=n yoziladi. xususiy yechimlarni ajratish uchun boshlang’ich shartlarni y(0)=u1, d(y)(0)=u2, (d@@2)(y)(0)=u3 va hokozalarni berish kerak bo’ladi. darajali qatorga yoyish turi series bo’ladi, shuning uchun keyinchalik bu qator bilan ishlash uchun uni convert(%,polynom) buyrug’i bilan polinom ajratish, so’ngra esa rhs(%) buyrug’i bilan olingan natijani o’ng tomonini ajratish kerak bo’ladi. misollar 1. y''(x)- y3(x)=ye - xcosx differensial tenglamaning umumiy yechimini 4-tartibli darajali qatorga yoyish ko’rinishida toping. yoyishni y(0)=1, y'(0)=0 boshlang’ich shartlarda amalga oshiring.. > restart; order:=4: de:=diff(y(x),x$2)-y(x)^3=exp(-x)*cos(x): > f:=dsolve(de,y(x),series); izoh: olingan yoyilmada d(y)(0) noldagi hosilani bildiradi: y'(0). xususiy yechimni topish …

4 / 7

$ordamida albatta ko’phad ko’rinishiga keltirish kerak. > convert(%,polynom): y2:=rhs(%): > p1:=plot(y1,x=-3..3,thickness=2,color=black): > p2:=plot(y2,x=-3..3, linestyle=3,thickness=2,color=blue): > with(plots): display(p1,p2); rasmda ko’rinib turibdiki, darajali qatorning aniq yechimiga yaqinlashishi taxminan - 1 restart; ordev=6: > eq:=diff(y(x),x$2)-x*sin(y(x))=sin(2*x): > cond:=y(0)=0, d(y)(0)=1: > de:=dsolve({eq,cond},y(x),numeric); de:=proc(rkf45_x)...end izoh: agar x o’zgaruvchining biror fiksirlangan qiymatida yechimni topish kerak bo’lsa , shu qiymat oldindan berilishi kerak, masalan, x=0.5 da quyidagi teriladi: > de(0.5); > with(plots): > odeplot(de,[x,y(x)],-10..10,thickness=2); endi koshi masalasining darajali qator ko’rinishida taqribiy yechimini topamiz va grafigini yasaymiz. > dsolve({eq, cond}, y(x), series) > convert(%, polynom):p:=rhs(%): > p1:=odeplot(de,[x,y(x)],-2..3, thickness=2,color=black): > p2:=plot(p,x=-2..3,thickness=2,linestyle=3,color=blue): > display(p1,p2); yechimning darajali qatorga yaqinlashuvi taxminan -1 restart; cond:=x(0)=1,y(0)=2: sys:=diff(x(t),t)=2*y(t)*sin(t)-x(t)-t,diff(y(t),t)=x(t): f:=dsolve({sys,cond},[x(t),y(t)],numeric): > with(plots): p1:=odeplot(f,[t,x(t)],-3..7, color=black, thicness=2,linestyle=3): p2:=odeplot(f,[t,y(t)],-3..7,color=green,thickness=2): > p3:=textplot([3.5,8,"x(t)"], font=[times, italic, 12]): > p4:=textplot([5,13,"y(t)"], font=[times, italic, 12]): > display(p1,p2,p3,p4); nazorat savollari 1. mapleda differensial tenglamalar qanday yechiladi? 2. yuqori tartibli differensial tenglamalar qanday yechiladi? 3. yechimning fundamental sistemasi qanday topiladi? 4. koshi masalasi qanday yechiladi? 5. differensial tenglamaning …$

5 / 7

image6.wmf image7.wmf image8.wmf image9.png image10.png image11.png image12.png image13.png image14.png image15.png image16.png image17.png image18.wmf image19.wmf image20.wmf image21.png image22.png image23.wmf image24.wmf image25.png image26.png image27.png image28.png image1.wmf image29.png image30.png image31.png image32.png image33.png image2.wmf image34.png image35.png image36.png image37.png image38.png image3.wmf image39.png image40.png image4.png = ( ) y x + - ( ) sin k x _c2 ( ) cos k x _c1 ( ) sin q x - + k 2 q 2 = ( ) y x + - ( ) sin k x _c2 ( ) cos k x _c1 1 2 ( ) cos k x x k := de = + + æ è ç ç ç ö ø ÷ ÷ ÷ ¶ ¶ 4 x 4 ( ) y x 2 æ è ç ç ç ö ø ÷ ÷ ÷ ¶ ¶ 2 x 2 ( ) y x ( ) y x 0 := de = + …

Want to read more?

Download all 7 pages for free via Telegram.

To'liq yuklab olish

About "ma’ruza 14. differensial tenglamalarni yechish"

ma’ruza 14. differensial tenglamalarni yechish. reja: 1. dsolve(t,f,options) buyrug’i. 2. yechimning fundamental (bazis) sistemasi. 3. differensial tenglamalarni sonli yechish. tayanch iboralar: differensial tenglama, koshi masalasi, chegaraviy masala. maple muhitida differensial tenglamalarni analitik yechish uchun dsolve(t,f,options) buyrug’i ishlatiladi, bu yerda t – differensial tenglama, f – noma’lum funksiya, options – parametrlar. parametrlar yechish metodlarini ko’rsatadi, masalan, jimlikda analitik yechim quyidagicha izlanadi: type=exact. differensial tenglamalarni tuzishda hosilalarni belgilash uchun diff buyrug’i ishlatiladi, masalan, y''+y=x differensial tenglama quyidagicha yoziladi: diff(y(x),x$2)+y(x)=x. differensial tenglamalarning umumiy yechimi ixtiyoriy o’zgarmasdan, ya’...

This file contains 7 pages in DOCX format (97,4 KB). To download "ma’ruza 14. differensial tenglamalarni yechish", click the Telegram button on the left.

Tags: ma’ruza 14. differensial tengla… DOCX 7 pages Free download Telegram